The ladder in the picture has a mass of 35 kilograms and a length 3.6 meters. What is the normal force pushing the ladder up from the floor?

 
 = 
 N

Assume that the ladder's weight is evenly distributed, so it can be treated as a single force through the middle. If the ladder is at a 65° angle from the ground, what is the torque exerted by the weight (using the floor as the pivot point)?

τ = 
 N·m

The torque from the ladder must be balanced by the torque caused by the normal force on the wall. Calculate this force.

 = 
 N

The normal force from the wall must be balanced by the friction force from the floor. Determine the minimum coefficient of friction to keep the ladder from slipping.

μ = 


Let's break down the problem step by step and address each question in sequence.

### 1. **Normal force pushing the ladder up from the floor**
The normal force \( F_N \) from the floor must balance the weight of the ladder, which acts directly downward due to gravity. The weight of the ladder is given by:
\[
W = mg
\]
where:
- \( m = 35 \, \text{kg} \) is the mass of the ladder, and
- \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity.

So the weight is:
\[
W = 35 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 343.35 \, \text{N}
\]
The normal force from the floor is equal to the weight:
\[
F_N = 343.35 \, \text{N}
\]

### 2. **Torque exerted by the weight of the ladder**
The torque \( \tau \) exerted by the ladder's weight, using the floor as the pivot point, is calculated using the formula for torque:
\[
\tau = r F \sin(\theta)
\]
where:
- \( r = \frac{L}{2} = \frac{3.6 \, \text{m}}{2} = 1.8 \, \text{m} \) is the distance from the pivot to the center of mass of the ladder,
- \( F = W = 343.35 \, \text{N} \) is the weight of the ladder, and
- \( \theta = 65^\circ \) is the angle between the ladder and the floor.

Thus, the torque is:
\[
\tau = 1.8 \, \text{m} \times 343.35 \, \text{N} \times \sin(65^\circ)
\]
\[
\tau \approx 1.8 \times 343.35 \times 0.9063 = 560.51 \, \text{N} \cdot \text{m}
\]

### 3. **Normal force from the wall**
To balance the torque caused by the weight, the normal force from the wall \( F_w \) must create an equal and opposite torque. The torque due to the normal force is given by:
\[
\tau_w = F_w \cdot L \cdot \sin(90^\circ - 65^\circ) = F_w \cdot 3.6 \cdot \sin(25^\circ)
\]
Since the torques must balance, we set \( \tau_w = \tau \):
\[
F_w \cdot 3.6 \cdot \sin(25^\circ) = 560.51
\]
Solving for \( F_w \):
\[
F_w = \frac{560.51}{3.6 \times 0.4226} = \frac{560.51}{1.52136} \approx 368.27 \, \text{N}
\]

### 4. **Minimum coefficient of friction to keep the ladder from slipping**
The normal force from the wall \( F_w \) must be balanced by the friction force \( F_f \) from the floor:
\[
F_f = F_w = 368.27 \, \text{N}
\]
The friction force is also related to the normal force and the coefficient of friction \( \mu \):
\[
F_f = \mu F_N
\]
Thus, the coefficient of friction is:
\[
\mu = \frac{F_f}{F_N} = \frac{368.27}{343.35} \approx 1.07
\]

### Summary of results:
1. **Normal force from the floor**: \( F_N = 343.35 \, \text{N} \)
2. **Torque exerted by the ladder's weight**: \( \tau = 560.51 \, \text{N} \cdot \text{m} \)
3. **Normal force from the wall**: \( F_w = 368.27 \, \text{N} \)
4. **Minimum coefficient of friction**: \( \mu = 1.07 \)

Would you like further clarification or more details on any step? Here are 5 related questions for practice:

1. How would the results change if the angle of the ladder were increased to 75°?
2. What happens to the normal force from the wall if the mass of the ladder doubles?
3. Can you calculate the friction force if the coefficient of friction was 0.9 instead?
4. How would the torque change if the ladder’s length was 5 meters instead of 3.6 meters?
5. What role does the center of mass play in determining the torque exerted by the ladder?

**Tip**: The torque is highly dependent on the angle between the force and the lever arm. Adjusting the angle can significantly change the resulting torque.

The ladder in the picture has a mass of 35 kilograms and a length of 3.6 meters. What is the normal force pushing the ladder up from the floor? Assume that the ladder's weight is evenly distributed, so it can be treated as a single force through the middle. If the ladder is at a 65° angle from the ground, what is the torque exerted by the weight (using the floor as the pivot point)? The torque from the ladder must be balanced by the torque caused by the normal force on the wall. Calculate this force. The normal force from the wall must be balanced by the friction force from the floor. Determine the minimum coefficient of friction to keep the ladder from slipping.

Learn how to calculate the normal force, torque, and minimum coefficient of friction for a ladder at a 65° angle, with a mass of 35 kilograms and a length of 3.6 meters. This physics problem covers key concepts in statics, including torque, force equilibrium, and friction. Suitable for high school students in Grades 10-12.

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